Factor Theorem

(ax-b) is a factor of \mathrm{f}(x) exactly when \mathrm{f}\!\left(\frac{b}{a}\right)=0, so find the value of x that makes each bracket zero and substitute it.

1. (2x-3)=0 when x=\frac32
2. \mathrm{f}\!\left(\frac32\right)=2\times\frac{27}{8}-9\times\frac94+\frac32+12=\frac{27}{4}-\frac{81}{4}+\frac{6}{4}+\frac{48}{4}=0
3. So (2x-3) is a factor.
4. Check the others are not: \mathrm{f}\!\left(-\frac32\right)=-\frac{33}{2}, \mathrm{f}(1)=6, \mathrm{f}(-4)=-264, none of which is 0.
5. Watch the sign: the factor (x+4) needs \mathrm{f}(-4)=0, not \mathrm{f}(4)=0.
6. Answer: (2x-3)

Write \mathrm{f}(x)=(x-4)(3x^2+bx+c) and match the x^2 term and the constant term to find b and c.

1. \mathrm{f}(x)=(x-4)(3x^2+bx+c): the 3x^2 is needed to make 3x^3
2. Constant term: -4c=4, so c=-1
3. x^2 term: b-12=-10, so b=2
4. Check the x term: c-4b=-1-8=-9 ✓
5. So \mathrm{f}(x)=(x-4)(3x^2+2x-1)
6. Factorise the quadratic: 3x^2+2x-1=(3x-1)(x+1)
7. Answer: (x-4)(3x-1)(x+1)

For part (a), (2x+1) is zero when x=-\frac12; after that, use your value of a and compare coefficients to find the quadratic factor.

Part (a)

1. (2x+1) is a factor, so \mathrm{f}\!\left(-\frac12\right)=0
2. 2\left(-\frac18\right)+a\left(\frac14\right)-6\left(-\frac12\right)-1=0
3. -\frac14+\frac{a}{4}+3-1=0
4. Multiply by 4: -1+a+8=0
5. Answer: a=-7

Part (b)

1. \mathrm{f}(x)=2x^3-7x^2-6x-1=(2x+1)(x^2+bx+c)
2. Constant term: c=-1
3. x^2 term: 2b+1=-7, so b=-4
4. Check the x term: 2c+b=-2-4=-6 ✓
5. Answer: x^2-4x-1

Part (c)

1. (2x+1)(x^2-4x-1)=0
2. 2x+1=0 gives x=-\frac12 (don't forget this one)
3. x^2-4x-1=0 does not factorise, so use the quadratic formula:
4. x=\dfrac{4\pm\sqrt{16+4}}{2}=\dfrac{4\pm\sqrt{20}}{2}=\dfrac{4\pm2\sqrt5}{2}=2\pm\sqrt5
5. Answer: x=-\frac12, x=2+\sqrt5, x=2-\sqrt5